Determining the type of conic from the parameterization $(t^2+t+1,t^2-t+1)$

Question

In the question the parametric coordinate of a conic is given by $(t^2+t+1,t^2-t+1)$

In questions which are similar to this I would just go on by writing "X=x coordinate expression" and similarly "Y = y coordinate expression" and then get value of "t" in terms of x and y after which i would equate both the values of "t" to get the locus expression from where determining conic type is easy.

But in this expression the parametric is in quadratic i am unable to figure out how to proceed.

Thanks in advance

Answer 1

Subtracting, we have $x-y=2t$, so $t=\dfrac{x-y}{2}$ .

Hence $$\begin{aligned}x=\frac{(x-y)^2}{4}+\frac{x-y}{2}+1 &\iff (x-y)^2+2(x-y)-4x+4=0 \\ &\iff (x-y+1)^2-4x+3=0\\ \end{aligned}$$

Now make a change of variables $Y:=x-y+1, X:=x-\frac34$ . Then $Y^2=4X$, which is a parabola.

Answer 2

$$x=\left(t+\frac12\right)^2+\frac34;\qquad y=\left(t-\frac12\right)^2+\frac34$$ Thus we have $t=\sqrt{x-\dfrac34}-\dfrac12$ so that

$$\left(t-\dfrac12\right)^2=\left(\sqrt{x-\dfrac34}-1\right)^2=x-\dfrac34+1-2\sqrt{x-\dfrac34} =x+\frac14-2\bigg(t+\frac12\bigg)=x-2t-\frac34$$

This gives $$y=x-2t$$ so that $(y-x)^2=4\left(x-\frac34\right)=4x-3$, or $$y^2+x^2-2xy-4x+3=0.$$ Following https://brilliant.org/wiki/conics-discriminant, we check the discriminant $$\Delta=\begin{vmatrix}1 &&-1&&-2\\-1&&1&&0\\-2 && 0&&3\end{vmatrix}=3+1(-3)-2(2)=3-3-4=-4.$$ Since $\Delta\neq0$, we check $B^2-4AC=h^2-ab=0$, which gives a parabola.